Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
#define int int64_t
#define MAX 3005
double dp[1503][MAX];
double a[MAX],s;
signed main()
{
	int n,i,j;
	cin>>n;
	for(i=1;i<=n;i++)
	cin>>a[i];
	for(i=0;i<=(n-1)/2;i++)
	{
		for(j=0;j<=n;j++)
		{
			if(j==0&&i==0)
			dp[i][j]=1;
			else
			{
			if(i>0)
			dp[i][j]=(dp[i-1][j-1])*a[j]+dp[i][j];
			dp[i][j]+=dp[i][j-1]*(1-a[j]);
		}
		}
	
		s=s+dp[i][n];
	}
	cout<<fixed<<setprecision(10)<<1-s<<"\n";
}